TITLE: How can I prove that $\int_{0}^{\infty }\frac{\log(1+x)}{x(1+x)}dx=\sum_{n=1}^{\infty }\frac{1}{n^2}$
QUESTION [5 upvotes]: How can I prove that $$\int_{0}^{\infty }\frac{\log(1+x)}{x(1+x)}dx=\sum_{n=1}^{\infty }\frac{1}{n^2}$$

REPLY [14 votes]: The substitution $u = \log(1+x)$ gives
$$I = \int_0^\infty \frac{\log(1+x)}{x(x+1)} dx = \int_0^\infty \frac{udu}{e^u-1}$$
Now by using $\frac{1}{1-x} = 1+x+x^2+\ldots$ with $x=e^{-u}$ we get
$$I = \sum_{k=0}^\infty\int_0^\infty  ue^{-(k+1)u}du$$
and by integration by parts it finally follows that
$$I = \sum_{k=0}^\infty\int_0^\infty  \frac{1}{k+1}e^{-(k+1)u}du = \sum_{k=0}^\infty\frac{1}{(k+1)^2}$$
Your integral can be generalized to
$$I(s) = \int_0^\infty \frac{\log^{s-1}(1+x)}{x(x+1)} dx = \int_0^\infty \frac{u^{s-1}du}{e^u-1} = \zeta(s)\Gamma(s)$$
where $\zeta(s)$ is the Riemann zeta function and $\Gamma(s)$ is the gamma function.